因为等差Sn=na1+d*n（n-1）/2 ;sn/sn-1式子与a1,n,d有关；而等比数列Sn=[a1*(1-q^n)]/(1-q)；设存在则有a2/a1=q,则a2=a1*q=a1+d,得出a1(q-1)=d①;a3=a1q²=a1+2d;a1(q²-1)=2d②将②/①则q=1,d=0; a4=a1q^3=...